For x^2 - 4x - 5 = 0 a student computes D = (-4)^2 - 4(1)(5) = 16 - 20 = -4 and concludes there are no real roots. What went wrong?

Question

Accepted Answer

c is -5, not 5, so D = 16 + 20 = 36 and the roots are 5 and -1. Here c = -5, so -4ac = -4(1)(-5) = +20 and D = 16 + 20 = 36. √D = 6, x = 4 ± 6/2 = 5 or -1. The formula is still b^2 - 4ac; only the sign of c was mishandled.

For $x^{2} - 4 x - 5 = 0$ a student computes $D = (- 4)^{2} - 4 (1) (5) = 16 - 20 = - 4$
and concludes there are no real roots. What went wrong?

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More Class 10 The Quadratic Formula practice

For x2−4x−5=0 a student computes D=(−4)2−4(1)(5)=16−20=−4and concludes there are no real roots. What went wrong?

Step-by-Step Solution

Master this chapter in 30 minutes a day.

More Class 10 The Quadratic Formula practice

For $x^{2} - 4 x - 5 = 0$ a student computes $D = (- 4)^{2} - 4 (1) (5) = 16 - 20 = - 4$
and concludes there are no real roots. What went wrong?